Part 8 · Chapter 36

Protective Relaying

A protective relay has to answer two questions from a few cycles of current and voltage — is this a fault, and is it mine? — and the whole subject is the collection of measuring principles that make the second question answerable without waiting to find out.

Electric Power Systems Prof. Mithun Mondal Reading time ≈ 55 min
i What you'll learn
  • Why reliability splits into dependability and security, and why improving one usually damages the other.
  • How the network is divided into overlapping zones at the CTs, and what remote and local backup each protect against.
  • What a CT's burden, accuracy limit factor and knee-point voltage mean, why saturation matters, and why the secondary must never be opened.
  • The IDMT characteristic \(t = \mathrm{TMS}\times 0.14/(M^{0.02}-1)\), the meaning of PSM and TMS, and how to grade a whole feeder from the far end backwards.
  • Why a directional element is needed on rings and parallel feeders, and what the maximum-torque angle does.
  • How percentage bias keeps a transformer differential stable through a CT-saturating through fault, and why second-harmonic restraint is not optional.
  • How a distance relay turns \(V/I\) into a point on the R–X diagram, and how the three zones are reached and timed.
  • What a numerical relay actually computes between the sample and the trip.
Section 36-1

The Four Qualities of Protection

Chapter 35 built a device that can interrupt twenty-five kiloamperes on command. This chapter builds the thing that gives the command. The two are useless apart, and the failure modes are different in kind: a breaker that cannot open is a mechanical failure with a single obvious remedy, while a relay that opens the wrong circuit has done something worse than nothing at all.

Protection is judged on four qualities, and they are in tension with one another. Understanding the tension is more valuable than memorising the list.

Selectivity, also called discrimination, means that only the faulty element is disconnected and everything healthy stays in service. A fault on one outgoing feeder of a substation must trip that feeder's breaker and not the incomer, even though the incomer sees exactly the same current. Selectivity is the quality that makes the whole subject difficult, because the current a relay measures is very nearly the same at every point of a radial chain — the information that distinguishes the relays is not in the magnitude of the current alone.

Speed matters for three separate reasons. Damage at the fault — burnt conductors, ruptured cable, a transformer winding destroyed rather than merely displaced — accumulates as \(\int i^2 dt\), the same integral that appeared in Section 35-9. The system's ability to stay in synchronism depends on clearing the fault before the accelerating area exceeds the available decelerating area, which is precisely the critical clearing time of Chapter 29. And the voltage depression during a fault stalls motors and drops out contactors across a wide area, so a fault cleared in three cycles is invisible to most customers while one cleared in a second is a regional event.

Sensitivity is the smallest fault the scheme will detect. A high-resistance earth fault on a resistance-earthed distribution system — a conductor lying on dry ground, a tree branch — may pass only a few amperes, far below any load current, and yet be lethal. Sensitivity is usually quoted as the minimum operating current or as the volt-amperes the relay draws at pickup.

Reliability is the quality most often stated loosely and it repays being split in two. Dependability is the probability that the scheme operates when it should. Security is the probability that it does not operate when it should not. Duplicating relays and connecting their trip contacts in parallel improves dependability and degrades security; requiring two relays to agree before tripping does the reverse. Every protection philosophy is a position on that trade, and transmission practice — where an unnecessary trip is expensive but a failure to trip is catastrophic — leans towards dependability, while generator and busbar protection leans towards security.

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Reliability is two quantities, not one
Dependability = it operates when required. Security = it stays put when not required. Redundancy in parallel buys dependability at the cost of security; redundancy in series does the opposite.

A fifth quality, stability, is really security applied to one specific case: a unit scheme is called stable if it does not operate for a fault outside its own zone, however large the through current. Section 36-7 is essentially a study of how to keep a differential relay stable.

Economy is a genuine constraint, not an afterthought. A 400 kV line justifies duplicate main protections on separate DC supplies with separate communication channels. A 415 V motor circuit justifies a fuse. Between them lies a continuum in which the cost of the protection is weighed against the cost of what it protects and the consequence of losing it — which is why the same fault type is protected five different ways at five different voltage levels.
Section 36-2

Zones, Primary Protection and Backup

Selectivity is achieved by dividing the network into zones of protection, one for each element that can be separately disconnected: each generator, each transformer, each busbar, each line, each large motor. A zone is bounded by the points at which current can be measured and interrupted — in practice, by circuit breakers with current transformers. The protection assigned to a zone is its primary protection, and it is required to trip every breaker on the zone boundary and no others.

The zones must overlap. If two adjacent zones merely met at a point, a fault exactly at that point would belong to neither and would be cleared by nobody. Overlap is arranged by taking the CTs for the two zones from opposite sides of the same circuit breaker, so that the breaker itself lies inside both zones. A fault in the small overlap region is then seen by both schemes and clears both — more of the system is lost than strictly necessary, but the region is a metre or two of busbar and the alternative is a blind spot.

G bus feeder 1 feeder 2 generator zone transformer zone busbar zone feeder zone zones overlap at the breaker: CTs taken from opposite sides so no point is unprotected
Every element has a zone, and adjacent zones overlap around the circuit breaker between them

Primary protection can fail, and it fails in more ways than one. The relay itself may be defective or wrongly set. The CT or VT feeding it may be faulty. The DC tripping supply may be lost. The trip coil may be open-circuit. And the breaker, having received a perfectly correct trip signal, may fail to open. Backup protection exists to cover all five, and it comes in two forms that cover different subsets.

Remote backup is provided by the protection of the next element upstream, delayed so that it acts only if the primary protection has not cleared the fault in its own time. It is completely independent — different relay, different CT, different DC supply, different breaker, different substation — and so it covers every failure mode including breaker failure. Its weakness is that it disconnects far more of the network than necessary and that its reach may be inadequate: an upstream relay may not see a fault at the far end of a long downstream feeder at all. The whole of Section 36-5 is an exercise in arranging remote backup.

Local backup sits in the same substation. A duplicate main protection using a different measuring principle and a separate CT core covers relay and CT failure without any time delay. Breaker-failure protection covers the remaining case: when a trip is issued, a timer of typically \(150\)–\(250\) ms starts, and if current is still flowing when it expires the scheme trips every other breaker on that busbar. Local backup is fast and precise but shares the substation's DC supply and building, so remote backup is retained as the last resort.

The relay–breaker split, recalled. Section 35-1 separated the decision from the action. Backup design is where that separation earns its keep: relay failure is covered by a second relay in the same substation, but breaker failure can only be covered by other breakers — which is why every protection scheme ultimately has an upstream relay watching it, and why the grading margin of Section 36-5 explicitly contains the breaker's own operating time.
Section 36-3

Instrument Transformers

No relay is connected to the primary system. Between the network and the relay sit the current transformer and the voltage transformer, whose job is to produce a scaled, isolated, low-level replica of the primary quantity. Everything a relay concludes is only as good as that replica, and most protection maloperations that are not setting errors are instrument-transformer errors.

A CT has its primary in series with the line, so the primary current is fixed by the network and is entirely indifferent to what the CT does. Its secondary works into a near short circuit — the relay and the leads. The equivalent circuit is the transformer of Chapter 3 with the burden on the secondary, and the source of all error is the magnetising branch: the current that flows into the magnetising impedance is current that does not reach the relay.

Where CT error comes from
\[ \frac{I_p}{N} = I_s + I_e, \qquad \varepsilon = \frac{I_e}{I_p/N} \]
\[ \text{with}\qquad E_s = I_s\big(R_{ct} + R_{lead} + R_{burden}\big) \quad\text{driving the flux that } I_e \text{ supplies} \]

Read the two lines together and the whole behaviour of a protection CT follows. The secondary emf that the core must generate is proportional to the secondary current and to the total resistance in the secondary loop — the CT's own winding resistance, the resistance of the run of pilot cable out to the relay panel, and the relay's own impedance. That emf demands flux; flux demands magnetising current; magnetising current is error. And because the core saturates, the demand can eventually be impossible to meet.

The total secondary impedance is the burden, quoted either in ohms or, more traditionally, as the volt-amperes it absorbs at rated secondary current: a \(5\) VA burden on a \(1\) A CT is \(5\) Ω, on a \(5\) A CT it is \(0.2\) Ω. The accuracy limit factor (ALF) states the multiple of rated primary current up to which the CT holds its stated accuracy at rated burden. A class \(5\text{P}20\) CT keeps its composite error within \(5\%\) up to twenty times rated current — but only if the burden it actually sees does not exceed the rated one. Reduce the burden and the effective ALF rises; exceed it and the CT saturates early.

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CT sizing
\[ V_k \;\ge\; \mathrm{ALF}\times I_{sn}\times\big(R_{ct} + R_{lead} + R_{burden}\big) \]

\(V_k\) is the knee-point voltage, defined as the secondary emf at which a \(10\%\) increase in applied voltage produces a \(50\%\) increase in magnetising current — the practical onset of saturation. The formula says something simple: the further the relay panel is from the switchyard, and the heavier the relay, the bigger the CT core has to be. Numerical relays with burdens of \(0.1\) Ω instead of \(3\) Ω have shrunk protection CTs considerably.

Saturation is worth understanding properly because it drives so much of protection design. If the core saturates, the secondary current is no longer a replica of the primary: it collapses for part of each cycle and reappears as a distorted spike, so its fundamental component is both reduced and phase-shifted. A relay measuring magnitude will under-read; a differential relay comparing two CTs, only one of which has saturated, will see a large spurious difference current. The DC offset of Chapter 21 makes it far worse, because a unidirectional component drives the flux steadily in one direction and can saturate a core that would handle the symmetrical current comfortably. Allowing fully for it requires a transient dimensioning factor of order \((1 + X/R)\) on the knee-point voltage, which for a transmission circuit means a core an order of magnitude larger than the steady-state calculation suggests. Modern practice accepts some saturation and designs the relay algorithm to ride through it.

One operating rule follows directly from the equivalent circuit and must never be broken: the secondary of a live CT must never be open-circuited. With \(I_s = 0\) the entire primary ampere-turns become magnetising ampere-turns. The core is driven deep into saturation, the flux waveform becomes a square wave whose transitions are extremely rapid, and \(N\,d\phi/dt\) at those transitions produces peaks of several kilovolts across the open terminals — enough to destroy the winding insulation and to kill whoever opened the circuit. Spare CT cores are short-circuited, and CT test blocks are made so that shorting happens before opening.

The voltage transformer is the easier device: primary across the system, secondary into a near open circuit, standard secondary \(110\) V line-to-line. Its errors come from the load current drawn by the burden through the winding impedance, so they grow with burden rather than with the measured quantity. Two constructions are used. The electromagnetic VT is a small wound transformer, accurate and well behaved, and it becomes expensive above about \(132\) kV because the primary insulation dominates the cost. The capacitor voltage transformer taps a point on a capacitive divider — usually the same stack that serves as a line coupling capacitor for carrier communication — and feeds an intermediate electromagnetic VT through a tuning reactor. It is far cheaper at EHV, but the divider stores energy, so after a sudden voltage collapse the secondary output takes a cycle or two to follow. Distance relays, which measure \(V/I\) in the instants after exactly such a collapse, must be designed for that transient or they will over-reach.

Finally, the way the three CTs are connected decides what the relay sees. Secondaries in star give the three phase currents and, from the star point, their sum. That sum is zero in balanced or purely phase-to-phase conditions and equals \(3I_0\) when earth current flows — the residual connection is therefore a direct hardware realisation of the zero-sequence component of Chapter 22, and a relay in the neutral lead is an earth-fault relay that is blind to load. Secondaries in delta produce differences of phase currents, which removes zero-sequence entirely and shifts the phase by \(30^\circ\): exactly what is needed to match across a star–delta transformer, as Section 36-7 requires.

Sequence components, in copper. Chapter 22 introduced \(I_0 = \tfrac13(I_a+I_b+I_c)\) as an algebraic device. A star-connected CT group computes it physically, with no relay involved, simply by letting the three secondaries share a return path. The residual connection is the reason earth-fault protection can be set at \(10\)–\(20\%\) of rated current while phase protection must be set above full load — the measured quantity is one that load current cannot produce.
Section 36-4

Overcurrent Relays and the IDMT Curve

The cheapest usable measurement is the magnitude of the current, and the oldest protective relay simply compares it with a threshold. Three time characteristics are built on that comparison.

An instantaneous relay operates with no intentional delay as soon as the current exceeds its setting. It gives the fastest possible clearance but can only be used where the setting can be placed above the highest fault current at the far end of the protected element, so that it never sees a fault beyond it. On a feeder with a substantial impedance of its own that is possible; on a short feeder it is not.

A definite-time relay operates after a fixed delay once the threshold is passed, independent of how large the current is. It grades easily, because each relay is simply given a longer delay than the one downstream, but it has an unattractive property: the most severe faults, closest to the source, are cleared most slowly.

The inverse-time relay fixes that. Its operating time falls as the current rises, so a heavy fault is cleared quickly and a marginal overload is given time to disappear on its own. On a radial feeder the fault current is naturally larger closer to the source, which means an upstream relay looking at a downstream fault sees a smaller multiple of its own setting than the downstream relay does, and therefore takes longer — grading falls out of the characteristic itself rather than having to be imposed. The practical relay is inverse definite minimum time (IDMT): inverse over the working range, flattening to a minimum time at very high currents where the curve would otherwise become unusably steep.

Two settings shape the relay. The plug setting or current setting fixes the threshold, expressed as a percentage of the rated secondary current; it decides whether the relay operates. The time multiplier setting (TMS), a number between about \(0.025\) and \(1.0\), scales the whole curve vertically; it decides when. The current is fed into the characteristic not in amperes but as the plug setting multiplier, the number of times the relay current exceeds its setting.

Plug setting multiplier
\[ \mathrm{PSM} \;=\; \frac{\text{relay current}}{\text{relay setting}} \;=\; \frac{I_f / (\text{CT ratio})}{\text{plug setting} \times I_{sn}} \;=\; \frac{I_f}{(\text{CT ratio})\times \text{plug setting}\times I_{sn}} \]

The denominator, written in primary amperes, is the primary pickup current, and it is the quantity to check first in any grading problem: it must sit above the maximum load the circuit will ever carry — including the cold-load pickup when a feeder is re-energised after an outage and every thermostatic load starts at once — and comfortably below the smallest fault current the relay must detect. A setting of about \(1.3\) times full load is typical for a feeder, while an earth-fault relay on the residual connection of Section 36-3 can be set at \(10\)–\(20\%\) because load current does not appear in the measured quantity at all.

IEC 60255 standardises four curve shapes, all of the same algebraic family.

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The IEC inverse-time characteristics
\[ t \;=\; \mathrm{TMS}\times\frac{k}{M^{\alpha}-1}, \qquad M = \mathrm{PSM} \]
\[ \text{SI: } \frac{0.14}{M^{0.02}-1} \quad\; \text{VI: } \frac{13.5}{M-1} \quad\; \text{EI: } \frac{80}{M^{2}-1} \quad\; \text{LTI: } \frac{120}{M-1} \]

SI is standard inverse, VI very inverse, EI extremely inverse, LTI long-time inverse. Note that \(t \to \infty\) as \(M \to 1\): a relay at exactly its setting never operates, which is why the pickup must be genuinely below the minimum fault current and not merely equal to it.

The choice among the four is a choice about how sharply time should fall with current. Standard inverse, with its exponent of \(0.02\), is very nearly flat — a hundredfold increase in current changes \(M^{0.02}\) only from about \(1.05\) to \(1.10\) — and it grades gently, which suits a chain of distribution relays. Extremely inverse falls as \(1/M^2\), which is the same shape as the thermal \(I^2t\) limit of a cable or the melting curve of a fuse, so it is the characteristic to use when an overcurrent relay must discriminate against downstream fuses or protect a machine on thermal grounds. Very inverse sits between the two and is preferred where the fault current changes strongly along a feeder, since a steeper curve extracts more time difference from a given current difference.

Section 36-5

Grading a Radial Feeder

Grading is the procedure that turns a set of identical relays into a selective scheme. The rule is simple to state: for any fault, the relay that should clear it must operate first, and every relay that also sees it must be slower by a definite margin. The margin is not arbitrary. It is built from four physical contributions.

ContributionTypical valueWhy it is there
Downstream breaker operating time60–100 msThe fault current persists until the arc is finally extinguished (Chapter 35)
Upstream relay overshoot30–50 ms electromechanical, ~10 ms numericalThe disc keeps turning, or the algorithm keeps integrating, after the current has gone
Combined CT and relay errors≈10% of the operating timeBoth relays have tolerances, and they can lie in opposite directions
Safety margin50–100 msSetting drift, temperature, and the fact that a failed grading is invisible until it matters
Total grading interval0.3–0.4 s (0.25–0.3 s all-numerical)

The procedure itself always runs from the load end backwards, because each relay's time is determined by the one below it:

1. Choose each relay's plug setting from load and minimum-fault considerations, working in primary amperes. 2. Give the most downstream relay the smallest TMS its curve allows for an acceptable time at the fault at the end of its own section. 3. Compute that relay's operating time for a fault at the end of its section. 4. Move one relay upstream. For the same fault, its required time is the downstream time plus the grading interval; compute the PSM it sees for that fault and solve for its TMS, rounding up to the nearest available step. 5. Compute the new relay's own operating time for a fault at the end of its section, and repeat from step 4.

Two details cause most of the errors students make. The grading check is always performed at the fault that is common to both relays — the fault at the boundary between them — not at each relay's own worst case. And the TMS must always be rounded up, never to the nearest value, because rounding down eats into a margin that was calculated to be only just sufficient.

fault current (A) t (s) 0.51.01.52.0 1k2k3k5k8k F1 (2 kA) F2 (3 kA) R3 400/1, TMS 0.25 R2 200/1, TMS 0.20 R1 100/1, TMS 0.20 0.37 s 0.35 s
Three graded standard-inverse relays: the margin is checked at the fault common to each pair

Example 3 works this through in full. The result exposes the characteristic weakness of the method, and it is worth naming in advance. As the grading marches back towards the source the operating times accumulate, so the relay protecting the most valuable equipment and seeing the largest fault current is the slowest in the chain. On a feeder with four or five sections the incomer may need well over a second, which is longer than the thermal limits of the switchgear and often longer than the critical clearing time of Chapter 29 permits.

Three remedies are used. A high-set instantaneous element is added to the upstream relay, set above the fault level at the downstream bus (typically \(1.3\) times it, to allow for DC offset and for the difference between the study and reality) so that a close-in fault is cleared in one cycle while faults beyond the next bus are left to the graded curve. A steeper characteristic — very or extremely inverse — extracts more time separation from the same current ratio and shortens the whole chain. Or the principle is abandoned: differential protection (Section 36-7) and distance protection (Section 36-8) both make selectivity a property of the measurement rather than of a stopwatch, and neither pays a time penalty for being upstream.

Section 36-6

Direction, and Earth Faults

Everything so far assumed a radial feeder, in which fault current can only flow one way. Two very common arrangements break that assumption. On parallel feeders, a fault on one of them is fed from the far bus back through the healthy feeder, so the healthy feeder's relay at the far end sees fault current and would trip a perfectly sound circuit. On a ring main, or any network with more than one source, current at any point can flow either way depending on where the fault is. Magnitude alone can no longer discriminate.

What distinguishes the two cases is the phase of the current relative to something that does not reverse. That reference — the polarising quantity — is a voltage. The classical induction-cup directional element develops a torque proportional to the product of the two quantities and the cosine of the angle between them:

Directional element
\[ T \;\propto\; V\,I\,\cos\big(\theta - \tau\big) \]
\[ \theta = \angle V - \angle I \;\;(\text{the system power-factor angle}), \qquad \tau = \text{relay characteristic angle (MTA)} \]

The torque is positive — the relay declares "forward" — over the half plane \(|\theta - \tau| < 90^\circ\), and the boundary is a straight line through the origin. The angle \(\tau\) at which torque is maximum is the maximum torque angle, and it is chosen so that the current a genuine forward fault produces sits near the middle of the operating half plane. Since fault current lags the driving voltage by the impedance angle of the network, \(\tau\) is set to \(30^\circ\) or \(45^\circ\) for phase-fault relays on distribution networks and higher on transmission circuits.

Choosing which voltage to use is the subtle part. Feeding the phase-A relay with \(V_a\) fails for the worst case, a close-up three-phase fault, because \(V_a\) collapses to nearly zero and the relay has no polarising quantity at all. The standard remedy is the \(90^\circ\) connection: the phase-A element is fed with \(I_a\) and the line voltage \(V_{bc}\), which is derived from the two healthy phases for any single-phase fault and which retains useful magnitude even for a three-phase fault a short distance away. The name comes from the fact that \(V_{bc}\) leads \(V_a\) by \(90^\circ\) at unity power factor. Numerical relays go further and hold a memory of the pre-fault voltage phasor for a few cycles, so that even a bolted three-phase fault at the relay terminals can be given a direction.

A directional relay never trips on direction alone. It is an overcurrent relay with a directional element in series: the current unit decides that a fault exists, the directional unit decides that it is in front. On a ring main this converts the ring into two independent radial chains for grading purposes — one traversing the ring clockwise, one anticlockwise — and each chain is graded exactly by the procedure of Section 36-5.

Earth-fault protection deserves its own measurement because the fault current can be so much smaller. The residual connection of Section 36-3 gives the relay \(I_a + I_b + I_c = 3I_0\), which is zero for load and for every phase-to-phase fault, so the setting is free of the load constraint and can be placed at \(10\)–\(20\%\) of rating. On a solidly earthed system that is enough. On a resistance-earthed or impedance-earthed system, where the designer has deliberately limited earth-fault current to a few hundred or even a few tens of amperes, a still more sensitive sensitive earth fault element is added, set at one or two per cent and fed from a core-balance CT — a single toroid through which all three phase conductors pass, so that the residual quantity is formed magnetically instead of by summing three separate CT errors.

Directional earth-fault relays need a polarising quantity that exists during an earth fault, and \(V_{bc}\) will not do. Two are used: the residual voltage \(3V_0\), obtained from an open-delta ("broken-delta") tertiary winding on the VTs, or the current in the transformer neutral, which by definition flows only towards a genuine earth fault. Both are direct hardware embodiments of the zero-sequence network of Chapter 23, and the choice between them turns on whether the substation has an earthed source of its own.

Why the sequence networks keep reappearing. Chapter 22 defined symmetrical components to make unbalanced fault algebra tractable. Protection uses them for a different reason entirely: each sequence quantity is a filter that is blind to some class of normal operation. Zero sequence ignores load, negative sequence ignores balanced conditions of any kind — which is why a negative-sequence element can detect a single-phase fault so weak that no phase current has changed noticeably.
Section 36-7

Differential Protection

Every scheme so far has been a graded scheme: it decides whether a fault is inside its zone by measuring how severe the fault is and how long everyone else has had to deal with it. A unit scheme decides directly, by comparing the boundaries of the zone with one another, and it is therefore both instantaneous and absolutely selective. The idea is Kirchhoff's current law applied to a piece of plant.

Bring the secondaries of the CTs at every boundary of the zone into a common circuit, arranged so that current circulating through the zone circulates round the pilot loop and produces no current in the relay. Then for any external condition — load, or a through fault of any magnitude — what enters equals what leaves and the relay sees nothing. For a fault inside the zone the two no longer balance, and the difference, which is the entire fault current fed into the zone, flows through the relay. This is the Merz–Price circulating current principle.

The ideal differential quantity
\[ I_{diff} = \big|\,\vec{I}_1 + \vec{I}_2\,\big| \;=\; 0 \;\;\text{externally}, \qquad I_{diff} = \big|\,\vec I_F\,\big| \;\;\text{internally} \]

In practice \(I_{diff}\) is never zero externally, and the reasons are worth listing because each one dictates a feature of the finished relay. The two CTs are not identical, so their ratio errors differ, and the difference grows with current. One CT may saturate while the other does not, which for a heavy through fault produces a spurious difference of many per unit — precisely when the relay must be most certain. A transformer's on-load tap changer alters the true ratio by up to \(\pm10\%\) away from the value the CTs were matched at. A transformer draws magnetising current that enters one side and leaves nowhere. And the pilot wires themselves have resistance.

The cure is percentage bias. Instead of comparing \(I_{diff}\) with a fixed threshold, compare it with a threshold that grows in proportion to how much current is passing through the zone. Define the through, or bias, current as the average of the terminal currents, and require

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Percentage-bias characteristic
\[ I_{diff} \;>\; I_s + k\,I_{bias}, \qquad I_{bias} = \tfrac12\big(|\vec I_1| + |\vec I_2|\big) \]

\(I_s\) is the minimum pickup — set above the magnetising current, typically \(0.1\)–\(0.3\) per unit — and \(k\) is the bias slope. Modern relays use two or three slopes: about \(20\)–\(30\%\) over the load and moderate-fault range, where only ratio errors and the tap changer contribute, rising to \(70\)–\(100\%\) above a few per unit, where CT saturation becomes the dominant error. The scheme is therefore most tolerant exactly where the errors are largest.

I_bias (pu) I_diff (pu) 14710 258 bias line: Is = 0.2, k = 30% then 70% OPERATE RESTRAIN internal fault (4.76, 9.52) → trip inrush (4.0, 8.0) → blocked by 2nd harmonic through fault (8.78, 0.45) → restrain
The bias line rises with through current, so the scheme is most tolerant where CT errors are largest

Applying the principle to a power transformer introduces three further complications that have nothing to do with errors and everything to do with the transformer being a transformer.

Ratio. The currents on the two sides differ by the transformer's turns ratio, so the CT ratios must be chosen to bring both secondary currents to the same value at rated load. An exact match is rarely available from standard CT ratios, and the residual mismatch is taken up by interposing CTs, by relay taps, or — in a numerical relay — by a scaling constant in software.

Phase shift. A star–delta transformer shifts phase by \(30^\circ\), so even perfectly matched magnitudes would leave a differential current of \(2\sin15^\circ = 0.52\) per unit. The classical compensation is to connect the CT secondaries in delta on the star side of the transformer and in star on the delta side, which introduces an equal and opposite \(30^\circ\) shift. The delta connection does a second job at the same time: it traps zero-sequence current, which is essential because an earth fault outside the zone on an earthed star winding drives zero-sequence current into that winding which has no counterpart on the delta side, and would otherwise look exactly like an internal fault.

Magnetising inrush. This is the difficult one. When a transformer is energised, the flux required is the integral of the applied voltage, and if the switching instant is unfavourable — the voltage passing through zero — the flux demanded in the first half cycle is twice the normal peak, on top of whatever remanence the core retains. The core saturates hard, and the current drawn to produce that flux can reach eight to twelve times rated current, decaying over several seconds. It flows into one side of the transformer and out of nowhere, so the differential relay sees the whole of it as difference current. A biased relay does not help: the bias current is only half the inrush, so the operating point sits well above the characteristic, as the figure shows.

What distinguishes inrush from a fault is its waveform. Because saturation clips the current for part of every cycle, inrush is strongly distorted and contains a large second-harmonic component — typically \(15\)–\(20\%\) of the fundamental, and never less than a few per cent — whereas a genuine internal fault current is close to sinusoidal, with second harmonic below about \(5\%\). Measuring the second harmonic and blocking the trip when it exceeds a set fraction is therefore both simple and decisive, and it is what every transformer differential relay has done for eighty years. A companion problem, overexcitation — a transformer run at high volts-per-hertz, which also saturates the core and also draws excess magnetising current — is distinguished instead by its fifth harmonic, and is blocked separately.

Unit protection has no reach and no time. A differential relay does not need to know the fault level, the source impedance, or what the neighbouring relays are doing; its zone is defined by where the CTs are bolted, and it can trip in one cycle without stealing anyone's grading margin. That is why every generator, transformer, busbar and cable of consequence has a differential scheme, and why the graded overcurrent behind it exists only as backup.
Section 36-8

Distance Protection and the R–X Diagram

Differential protection needs a communication channel between the ends of the zone, and on a two-hundred-kilometre transmission line that channel is a substantial part of the cost and a new way for the scheme to fail. Graded overcurrent, on the other hand, breaks down on transmission circuits for a reason worth stating precisely: the relay setting depends on the fault level, the fault level depends on how many generators happen to be running, and on a heavily loaded line the minimum fault current at the remote end may not comfortably exceed the maximum load current. A measurement is needed whose answer does not move when the generation pattern moves.

Distance protection supplies it. Measure both the voltage and the current at the relay point and form their ratio. For a bolted fault at a distance \(\ell\) along a line of impedance \(z\) ohms per kilometre, the voltage at the relay is the drop along that much line, so

What a distance relay measures
\[ Z_{relay} = \frac{V_{relay}}{I_{relay}} = \frac{I\,z\,\ell}{I} = z\,\ell \]

The current cancels. The measured impedance is a property of the line alone, proportional to the distance to the fault, and entirely independent of the source strength behind the relay. That single cancellation is the whole reason distance protection exists.

The measurement is a complex number, so it is natural to plot it on the R–X diagram, with resistance along the horizontal axis and reactance along the vertical. On that plane a healthy line appears as a straight line from the origin at the line's impedance angle — about \(75^\circ\)–\(85^\circ\) for an overhead transmission line — with length proportional to the line length. A fault at a given point is a point on that line. Load, by contrast, appears at a much larger radius and a much smaller angle, since a load impedance is large and near unity power factor. A relay characteristic is then simply a region of the R–X plane: the relay trips when the measured point falls inside it.

Four regions are in common use. The plain impedance characteristic is a circle centred on the origin — it responds to \(|Z|\) alone, which makes it non-directional and gives it very poor tolerance of load. The reactance characteristic is a horizontal line, tripping when \(X\) is less than a set value; because it ignores \(R\) entirely it is completely tolerant of arc resistance, but for the same reason it will happily trip on load and must be supervised by another element. The mho characteristic is a circle passing through the origin with its diameter along the line angle: it is inherently directional (the origin is on its boundary, so nothing behind the relay is enclosed), it is compact in the direction where load lies, and it is the workhorse of electromechanical and static distance relays. Numerical relays are not restricted to shapes a rotating disc can produce and generally use a quadrilateral, whose reactive reach and resistive reach are set independently — the best of the reactance and mho characteristics at once.

R (Ω) X (Ω) Zone 3 — 26.7 Ω, 0.8 s Zone 2 — 14.7 Ω, 0.35 s Zone 1 — 8.5 Ω, instant bus B adjacent lines A 75° load: 63.5 Ω at 26° — far outside every zone
Three mho zones in secondary ohms, the protected line, and the load impedance they must not enclose

A relay does not measure primary ohms. The CT scales the current down by \(N_{CT}\) and the VT scales the voltage down by \(N_{VT}\), so the impedance the relay sees is scaled by their ratio, and every setting must be converted before it is entered:

Primary to secondary ohms
\[ Z_{sec} = Z_{pri}\times\frac{N_{CT}}{N_{VT}} \]

The three zones are the practical realisation of the scheme, and each one exists for a distinct reason.

Zone 1 is set to \(80\%\) of the protected line and trips instantaneously. It stops short of the remote bus deliberately. The measured impedance carries errors from the CT, the VT, the relay itself and the line impedance data, and if those errors conspired while the reach was set to \(100\%\), the relay would trip instantaneously for a fault on the next line — an unselective operation that no time delay could correct, because there is none. The \(20\%\) shortfall is the price of never over-reaching.

Zone 2 covers the last \(20\%\) that zone 1 gave up, and therefore must reach beyond the remote bus — at least \(120\%\) of the line. It must not, however, reach past the zone 1 of the shortest line leaving the remote bus, or the two would trip together with no discrimination. The usual setting is the protected line plus \(50\%\) of the shortest adjacent line, and a delay of \(0.3\)–\(0.4\) s lets the remote relay's zone 1 act first for faults it can see.

Zone 3 is remote backup: the protected line plus \(120\%\) of the longest adjacent line, with a delay of \(0.8\)–\(1.0\) s. It is the distance relay's contribution to the backup structure of Section 36-2, covering failure of the remote substation's protection or breaker.

Three effects complicate every setting. Arc resistance adds a purely resistive component to the measured impedance, moving the point to the right on the R–X diagram and possibly out of a mho circle for a fault near the reach limit — the argument for a quadrilateral characteristic with generous resistive reach. Infeed from a source at the remote bus means that the current through the far section is larger than the current the relay measures, so the relay under-reads the distance and its zone 2 covers less of the adjacent line than intended; Example 6 works out how much. And load encroachment is the risk that a heavily loaded line, particularly during the low-voltage conditions of an emerging system disturbance, presents an impedance small enough to enter zone 3. The power swings of Chapter 28 do exactly this — the apparent impedance during a swing travels across the R–X plane and can pass through the zone 3 circle — which is why every transmission distance relay carries a power-swing blocking element that distinguishes the slow, continuous movement of a swing from the instantaneous jump of a fault.

A complete distance relay is not one measuring element but six: three phase-fault loops (A–B, B–C, C–A) and three earth-fault loops (A–E, B–E, C–E), because the impedance seen depends on which conductors the fault involves. The earth loops need one further correction, since the return path is through the earth and the zero-sequence impedance of Chapter 23 is not the positive-sequence impedance. The loop is compensated by the residual compensation factor

Earth-loop compensation
\[ Z_{A\text{-}E} = \frac{V_a}{I_a + k_0\,3I_0}, \qquad k_0 = \frac{Z_0 - Z_1}{3\,Z_1} \]

so that the earth loop measures the same positive-sequence ohms per kilometre as the phase loops and one set of zone reaches serves both.

Section 36-9

Inside a Numerical Relay

Every characteristic in this chapter was invented for a relay made of iron and copper: a rotating induction disc with a spring restraint gave the inverse-time curve, an induction cup gave the \(VI\cos(\theta-\tau)\) torque, a pair of coils in opposition gave the percentage bias, and a balanced beam gave the mho circle. Since the 1990s all of it has been done in software, and the shift is worth understanding because it changes what is possible without changing any of the physics.

The signal chain is short. The CT and VT secondaries pass through small internal transformers and burden resistors into an anti-aliasing filter, a low-pass filter with a corner well below half the sampling rate, which is not optional: a harmonic above the Nyquist frequency would otherwise be folded down and appear as a spurious low-frequency component in the very quantity the relay is protecting on. The filtered signals are sampled simultaneously — simultaneity across channels matters, because every phase comparison in the chapter depends on it — at typically \(16\) to \(96\) samples per cycle, and digitised to \(12\) or \(16\) bits.

From the samples the relay extracts the fundamental phasor with a one-cycle discrete Fourier transform:

Full-cycle DFT phasor estimation
\[ X_c = \frac{2}{N}\sum_{k=0}^{N-1} x_k \cos\!\frac{2\pi k}{N}, \qquad X_s = \frac{2}{N}\sum_{k=0}^{N-1} x_k \sin\!\frac{2\pi k}{N} \]
\[ |X| = \sqrt{X_c^2 + X_s^2}\Big/\sqrt2, \qquad \angle X = \tan^{-1}\!\frac{-X_s}{X_c} \]

The full-cycle DFT rejects every harmonic exactly, which is why the second- and fifth-harmonic restraints of Section 36-7 come free — the same transform that produces the fundamental produces the harmonics as a by-product. It does not reject the DC offset, whose exponential decay leaks into the fundamental estimate; a mimic filter, a digital implementation of \(1 + \tau\,d/dt\) matched to the circuit's \(L/R\), removes it before the transform. The window is the reason a numerical relay cannot be arbitrarily fast: a one-cycle window means the estimate is not fully converged until one cycle after the fault, and typical zone 1 distance operating times of one to one and a half cycles are set by that, not by the processor.

With phasors in hand every characteristic in this chapter is a few lines of arithmetic. The IDMT curve is evaluated from the formula rather than approximated by a disc. The mho circle becomes a phase comparison between two derived phasors. The bias characteristic is an inequality. And because the characteristic is now a piece of code, shapes that no mechanism could produce — the quadrilateral, adaptive reaches, settings groups that change automatically when the network configuration changes — become ordinary.

Three consequences reach beyond the measurement itself. Self-monitoring changes the reliability arithmetic of Section 36-1 fundamentally: an electromechanical relay that failed sat silently until the next test, which might be years away, whereas a numerical relay continuously checks its memory, its analogue chain and its output circuits and raises an alarm within seconds. Disturbance recording means every operation is accompanied by a sampled record of what the relay saw, so post-fault analysis stops being guesswork. And communication, standardised by IEC 61850, allows relays to exchange trip and interlock signals as GOOSE messages over a substation Ethernet network instead of over hard-wired copper — which makes breaker-failure schemes, busbar blocking schemes and adaptive protection cheap enough to use everywhere rather than only on the largest circuits.

🔑
What changed and what did not
The measuring principles — magnitude, direction, difference, impedance — are exactly those of the electromechanical era. What numerical technology changed is the cost of implementing them, the freedom to choose the characteristic's shape, and the ability of the relay to tell you it has failed.

This is why the settings calculations of Sections 36-4 to 36-8 are still done the same way, and why a grading study performed in 1960 would still be recognised, and largely still correct, today.

Section 36-10

Worked Examples

1 Sizing a protection CT

Problem. A \(400/1\) A class \(5\text{P}20\) current transformer has a secondary winding resistance of \(4\) Ω and is connected to a relay panel by a pilot loop of \(1\) Ω. Find the knee-point voltage required (a) for a numerical relay of \(0.1\) Ω burden and (b) for an electromechanical relay of \(3\) VA. If the CT actually supplied has \(V_k = 102\) V, at what primary current does it saturate in case (b)? Comment on the effect of a network \(X/R\) of \(10\).

Solution. The relay burden in ohms is the VA rating divided by the square of the rated secondary current, so \(3\) VA at \(1\) A is \(3\) Ω while \(0.1\) Ω is simply \(0.1\) Ω.

Required knee-point voltage
\[ \text{(a)}\quad V_k \ge 20 \times 1 \times (4 + 1 + 0.1) = 20 \times 5.1 = 102\ \text{V} \]
\[ \text{(b)}\quad V_k \ge 20 \times 1 \times (4 + 1 + 3) = 20 \times 8 = 160\ \text{V} \]

Replacing the electromechanical relay by a numerical one has cut the required core by more than a third, which is the whole reason modern protection CTs are so much smaller than their predecessors.

With the \(102\) V CT driving the \(8\) Ω loop of case (b), the accuracy limit is reached when the secondary emf reaches the knee point:

Effective accuracy limit factor
\[ \mathrm{ALF}_{eff} = \frac{V_k}{I_{sn}\,(R_{ct}+R_{lead}+R_{burden})} = \frac{102}{1\times 8} = 12.75 \]
\[ I_{p,\,sat} = 12.75 \times 400 = 5100\ \text{A} \]

The nameplate says \(5\text{P}20\), but in this installation the CT holds accuracy only to \(12.75\) times rating. An \(8\) kA fault — a perfectly ordinary value — saturates it, and the relay under-reads.

The DC offset makes matters far worse. Allowing fully for a network \(X/R\) of \(10\) requires a transient dimensioning factor of \(K_{td} = 1 + X/R = 11\) applied to the actual fault current rather than to the rated current. For the \(8\) kA fault, \(20\) A secondary through \(5.1\) Ω:

Fully transient-rated core
\[ V_k \ge K_{td}\,I_{s}\,(R_{ct}+R_b) = 11 \times 20 \times 5.1 = 1122\ \text{V} \]

Eleven times the steady-state requirement. Cores of that size exist (classes TPX, TPY, TPZ) and are specified for busbar and generator differential schemes, but they are large and expensive. For everything else the modern approach is to accept that the CT will saturate for part of the first few cycles and to make the relay algorithm immune to it — which is a thing a numerical relay can do and an induction disc cannot.

2 PSM, TMS and the choice of curve

Problem. A standard-inverse relay of \(5\) A rating is fed from a \(400/5\) CT, has a plug setting of \(125\%\) and a TMS of \(0.3\). A fault of \(4000\) A occurs. Find the operating time. Repeat for very inverse and extremely inverse characteristics with the same settings, and find the effect of raising the plug setting to \(150\%\).

Solution. The plug setting multiplier can be formed in secondary or primary terms; both must agree.

Plug setting multiplier
\[ \text{relay current} = 4000\times\frac{5}{400} = 50\ \text{A}, \qquad \text{setting} = 1.25\times 5 = 6.25\ \text{A} \]
\[ \mathrm{PSM} = \frac{50}{6.25} = 8 \qquad\text{or}\qquad \mathrm{PSM} = \frac{4000}{400\times 1.25} = \frac{4000}{500} = 8 \;\;\text{(primary pickup 500 A)} \]
Operating times
\[ \text{SI:}\quad t = 0.3\times\frac{0.14}{8^{0.02}-1} = 0.3\times\frac{0.14}{0.04247} = 0.3 \times 3.297 = 0.989\ \text{s} \]
\[ \text{VI:}\quad t = 0.3\times\frac{13.5}{8-1} = 0.3\times 1.929 = 0.579\ \text{s} \]
\[ \text{EI:}\quad t = 0.3\times\frac{80}{8^{2}-1} = 0.3\times\frac{80}{63} = 0.381\ \text{s} \]

At the same PSM and TMS the steeper curves are much faster, which is why very and extremely inverse are chosen where fault current is high relative to the setting. The gain is not free: at low multiples they are correspondingly slower, so a marginal fault at the far end of a long feeder may take an unacceptable time.

Raising the plug setting to \(150\%\) moves the primary pickup to \(600\) A and the multiplier to \(4000/600 = 6.67\):

Effect of a higher plug setting
\[ t = 0.3\times\frac{0.14}{(6.67)^{0.02}-1} = 0.3\times\frac{0.14}{0.03867} = 0.3\times 3.620 = 1.086\ \text{s} \]

The relay has become \(10\%\) slower for the same fault. Plug setting and time multiplier are not independent knobs: raising the current setting to gain margin over load also lengthens every operating time on the curve, and a grading study must be redone if either is changed.

3 Grading a three-relay radial feeder

Problem. An \(11\) kV radial system runs Source → bus A → bus B → bus C → load. Relay R3 at A protects section A–B, relay R2 at B protects B–C, and relay R1 at C protects the outgoing feeder. All three are standard inverse. CT ratios and maximum loads are \(400/1\) with \(320\) A at A, \(200/1\) with \(160\) A at B, \(100/1\) with \(80\) A at C. Three-phase fault currents are \(8000\) A at bus A, \(5000\) A at bus B, \(3000\) A at bus C, and \(2000\) A at the far end of the outgoing feeder. Grade the three relays with a margin of \(0.3\) s, using TMS steps of \(0.05\).

Solution — plug settings. Take \(100\%\) plug on each relay, giving primary pickups of \(100\), \(200\) and \(400\) A. Each is \(1.25\) times the maximum load on its circuit, and each is far below the smallest fault current it must see (the smallest is \(2000\) A, giving R3 a multiple of \(5\)). Both constraints are satisfied.

R1, the most downstream relay. Start at the lowest useful multiplier setting, TMS \(= 0.1\), and evaluate at the feeder-end fault F1 \(=2000\) A:

R1 at F1
\[ \mathrm{PSM} = \frac{2000}{100} = 20, \qquad 20^{0.02} = 1.0617 \]
\[ t_{R1} = 0.1\times\frac{0.14}{0.0617} = 0.1\times 2.267 = 0.227\ \text{s} \]

R2, graded against R1 at the same fault F1. R2 must take \(0.227 + 0.30 = 0.527\) s for that fault, and it sees a smaller multiple because its CT and pickup are twice as large:

R2 graded at F1
\[ \mathrm{PSM} = \frac{2000}{200} = 10, \qquad \frac{0.14}{10^{0.02}-1} = \frac{0.14}{0.04713} = 2.971 \]
\[ \mathrm{TMS} = \frac{0.527}{2.971} = 0.177 \;\longrightarrow\; \text{adopt } \mathrm{TMS} = 0.20 \]
\[ t_{R2}\big|_{F1} = 0.20\times 2.971 = 0.594\ \text{s}, \qquad \text{margin} = 0.594-0.227 = 0.367\ \text{s} \;\ge 0.30 \]

R2's own duty. Its section ends at bus C, where the fault F2 is \(3000\) A:

R2 at F2
\[ \mathrm{PSM} = \frac{3000}{200} = 15, \qquad \frac{0.14}{15^{0.02}-1} = \frac{0.14}{0.05565} = 2.516 \]
\[ t_{R2}\big|_{F2} = 0.20\times 2.516 = 0.503\ \text{s} \]

R3, graded against R2 at F2. Required time \(= 0.503 + 0.30 = 0.803\) s:

R3 graded at F2
\[ \mathrm{PSM} = \frac{3000}{400} = 7.5, \qquad \frac{0.14}{(7.5)^{0.02}-1} = \frac{0.14}{0.04112} = 3.405 \]
\[ \mathrm{TMS} = \frac{0.803}{3.405} = 0.236 \;\longrightarrow\; \text{adopt } \mathrm{TMS} = 0.25 \]
\[ t_{R3}\big|_{F2} = 0.25\times 3.405 = 0.851\ \text{s}, \qquad \text{margin} = 0.851-0.503 = 0.348\ \text{s} \;\ge 0.30 \]

The settings are fixed. Tabulating every relay against every fault gives the complete picture:

FaultCurrentR1 (100/1, TMS 0.10)R2 (200/1, TMS 0.20)R3 (400/1, TMS 0.25)
F1 — feeder end2000 APSM 20 → 0.227 sPSM 10 → 0.594 sPSM 5 → 1.070 s
F2 — bus C3000 APSM 15 → 0.503 sPSM 7.5 → 0.851 s
F3 — bus B5000 APSM 12.5 → 0.676 s
F4 — bus A8000 APSM 20 → 0.567 s

Bold entries are the primary clearance; every other entry is graded backup, and each sits at least \(0.3\) s above the relay it backs up. R3's time at F1, \(1.070\) s, is \(0.476\) s above R2 — comfortably more than the minimum, which is what happens whenever a relay backs up a fault two sections away.

Now read the last two rows. The most severe fault in the whole system, \(8000\) A right at the source bus, is cleared in \(0.567\) s — the slowest primary clearance anywhere except the backup times, and it occurs where the fault energy is greatest and where the switchgear and the generator transformer are most valuable. This is the structural weakness of graded overcurrent described in Section 36-5. The remedy here is an instantaneous high-set element on R3, set above \(1.3 \times 5000 = 6500\) A, which clears the \(8000\) A bus-A fault in one cycle while leaving every fault at or beyond bus B to the graded curve, where the grading just computed remains untouched.

4 Biased differential protection of a transformer

Problem. A \(20\) MVA, \(132/33\) kV, YNd11 transformer of \(10.5\%\) impedance is protected by a biased differential relay with \(I_s = 0.2\) pu, a \(30\%\) slope up to \(4\) pu of bias and \(70\%\) beyond, the first breakpoint being at \(1\) pu. HV CTs are \(100/1\) and LV CTs are \(400/1\). (a) Verify the CT ratio match. (b) Check the relay for a three-phase fault at the LV terminals inside the zone. (c) Check its stability for a \(9\) pu through fault during which one CT reads \(5\%\) low. (d) Check the effect of the tap changer at \(+10\%\). (e) Check what happens on energisation.

Solution (a). The rated currents are

Rated currents and CT secondary currents
\[ I_{HV} = \frac{20\times10^{6}}{\sqrt3\,(132\times10^{3})} = 87.5\ \text{A}, \qquad I_{LV} = \frac{20\times10^{6}}{\sqrt3\,(33\times10^{3})} = 349.9\ \text{A} \]
\[ \frac{87.5}{100} = 0.875\ \text{A}, \qquad \frac{349.9}{400} = 0.875\ \text{A} \]

The two secondary currents are equal at rated load, so the ratio match is exact and \(1\) per unit may be taken as \(0.875\) A on either side. The \(30^\circ\) shift of the Yd11 vector group is removed by connecting the HV (star-side) CTs in delta and the LV (delta-side) CTs in star, which also strips the zero-sequence current that an external earth fault would otherwise inject into the HV winding alone.

Solution (b) — internal fault. With no source on the LV side, all the fault current enters from the HV side and none leaves:

Internal three-phase fault at the LV terminals
\[ I_F = \frac{1}{0.105} = 9.52\ \text{pu}, \qquad I_{diff} = 9.52\ \text{pu}, \qquad I_{bias} = \tfrac12(9.52 + 0) = 4.76\ \text{pu} \]
\[ \text{threshold} = 0.2 + 0.30(4-1) + 0.70(4.76-4) = 0.2+0.90+0.53 = 1.63\ \text{pu} \]

\(9.52 \gg 1.63\): the relay operates, with a factor of nearly six in hand.

Solution (c) — through fault with CT error. A fault just outside the LV CT drives \(9\) pu through the transformer. If the LV CT partially saturates and delivers only \(95\%\) of the true current:

Stability check
\[ I_{diff} = 9.00 - 8.55 = 0.45\ \text{pu}, \qquad I_{bias} = \tfrac12(9.00+8.55) = 8.78\ \text{pu} \]
\[ \text{threshold} = 0.2 + 0.90 + 0.70(8.78-4) = 1.10 + 3.34 = 4.44\ \text{pu} \]

\(0.45 \ll 4.44\): the relay restrains, with a factor of ten of margin. That margin is exactly what the steep second slope was introduced to provide, and it is the reason the characteristic is not a single straight line.

Solution (d) — tap changer. At the \(+10\%\) tap the true ratio differs from the value the CTs were matched at, so the same \(9\) pu through fault produces a genuine \(10\%\) mismatch:

Tap-changer mismatch
\[ I_{diff} = 0.10 \times 9 = 0.90\ \text{pu}, \qquad I_{bias} \approx 8.55\ \text{pu}, \qquad \text{threshold} = 1.10+0.70(4.55) = 4.29\ \text{pu} \]

Still stable. Tap-changer mismatch and CT error together would give \(1.35\) pu against a threshold of \(4.3\) pu, so the setting tolerates both at once.

Solution (e) — energisation. Inrush of \(8\) pu enters the HV winding and leaves nowhere:

Magnetising inrush
\[ I_{diff} = 8.0\ \text{pu}, \qquad I_{bias} = 4.0\ \text{pu}, \qquad \text{threshold} = 0.2 + 0.90 = 1.10\ \text{pu} \]

The operating point lies far inside the trip region, and the bias characteristic provides no protection whatever against it — inrush is indistinguishable from an internal fault on magnitude alone, because on magnitude alone it is one. Discrimination comes from the waveform: the saturated inrush current typically carries \(15\)–\(20\%\) second harmonic against under \(5\%\) for a fault, so a second-harmonic restraint set at \(15\%\) blocks the trip for the second or two the inrush lasts. Without that element the transformer could not be switched in.

5 Setting the three zones of a distance relay

Problem. A \(132\) kV line A–B is \(80\) km long with \(z = 0.4\ \Omega\)/km at \(75^\circ\). Two lines leave bus B: B–C of \(60\) km and B–D of \(100\) km, both with the same \(z\). The relay at A is fed by a \(400/1\) CT and a \(132\,000/110\) VT. Set the three zones in primary and secondary ohms with their time delays, and verify that zone 3 does not encroach on load when the line carries \(400\) A at \(0.9\) power factor with the voltage depressed to \(0.9\) pu.

Solution. The line impedances are \(Z_{AB} = 32\) Ω, \(Z_{BC} = 24\) Ω, \(Z_{BD} = 40\) Ω, all at \(75^\circ\). The VT ratio is \(132\,000/110 = 1200\), so

Primary-to-secondary conversion
\[ Z_{sec} = Z_{pri}\times\frac{N_{CT}}{N_{VT}} = Z_{pri}\times\frac{400}{1200} = \frac{Z_{pri}}{3} \]
Zone reaches
\[ \text{Zone 1} = 0.8\,Z_{AB} = 0.8(32) = 25.6\ \Omega \;\to\; 8.53\ \Omega\ \text{sec}, \qquad t_1 = 0 \]
\[ \text{Zone 2} = Z_{AB} + 0.5\,Z_{BC} = 32 + 12 = 44\ \Omega \;\to\; 14.67\ \Omega\ \text{sec}, \qquad t_2 = 0.35\ \text{s} \]
\[ \text{Zone 3} = Z_{AB} + 1.2\,Z_{BD} = 32 + 48 = 80\ \Omega \;\to\; 26.67\ \Omega\ \text{sec}, \qquad t_3 = 0.8\ \text{s} \]

Zone 2 is checked against its lower bound: it must reach at least \(1.2 \times 32 = 38.4\) Ω to guarantee cover of the last \(20\%\) of the line that zone 1 gave up, and \(44 > 38.4\). It is also checked against its upper bound: it must not reach past the zone 1 of the shortest adjacent line, which starts at \(32\) Ω and ends at \(32 + 0.8(24) = 51.2\) Ω, and \(44 < 51.2\). Both constraints are met.

Load check. The apparent impedance under the stated load is

Load impedance and the mho reach along it
\[ Z_{load} = \frac{0.9\times132\,000/\sqrt3}{400} = \frac{68\,590}{400} = 171.5\ \Omega\ \text{pri} = 57.2\ \Omega\ \text{sec}, \quad \angle 25.8^\circ \]
\[ \text{mho reach at } 25.8^\circ = 26.67\cos(75^\circ - 25.8^\circ) = 26.67\times 0.653 = 17.4\ \Omega\ \text{sec} \]

The load point sits at \(57.2\) Ω where the zone 3 circle extends only to \(17.4\) Ω — a margin of \(3.3\) times. The load is safely outside. The calculation also shows why the mho shape is used: a plain impedance circle of the same reach would extend the full \(26.67\) Ω in every direction, cutting the margin by more than a third, and on a more heavily loaded line it would not clear at all.

6 How infeed shortens zone 2

Problem. Continue Example 5. A second source is connected at bus B and, for a fault on line B–C, contributes twice the current that flows through the relay at A. Find how far along B–C the zone 2 setting of \(44\) Ω actually reaches, and state what setting would be needed to restore the intended \(50\%\) coverage.

Solution. Let the fault be \(d\) km along B–C, so \(Z_{Bx} = 0.4d\). The relay at A measures the ratio of its own voltage to its own current, but the drop along the faulted section is produced by the total current \(I_A + I_B\):

Apparent impedance with infeed
\[ V_A = I_A Z_{AB} + (I_A + I_B)\,Z_{Bx} \]
\[ Z_{app} = \frac{V_A}{I_A} = Z_{AB} + \left(1 + \frac{I_B}{I_A}\right)Z_{Bx} = Z_{AB} + K\,Z_{Bx} \]

With \(I_B/I_A = 2\) the infeed factor is \(K = 3\). Setting \(Z_{app}\) equal to the zone 2 reach:

Actual reach
\[ 32 + 3(0.4d) = 44 \;\Longrightarrow\; 1.2d = 12 \;\Longrightarrow\; d = 10\ \text{km} \]
\[ \text{without infeed:}\quad 32 + 0.4d = 44 \;\Longrightarrow\; d = 30\ \text{km} \]

Zone 2 has shrunk from \(50\%\) of line B–C to \(16.7\%\). Infeed always causes under-reach, never over-reach, because the extra current makes the fault look further away than it is — a reassuring direction of error for zone 1, which is why zone 1's \(80\%\) setting is not affected by infeed at all (there is no infeed within the protected line).

Restoring \(30\) km of coverage would need

The setting that would be required
\[ Z_2 = 32 + 3(0.4\times30) = 32 + 36 = 68\ \Omega \]

and this is exactly what must not be done. If the source at B is switched out, \(K\) returns to \(1\) and a \(68\) Ω zone 2 reaches \(90\) km along a \(60\) km line — past bus C and well into the circuits beyond it, tripping in \(0.35\) s for faults two lines away. The setting must be chosen for the weakest infeed condition and the reduced coverage accepted. Securing the last \(20\%\) of line A–B is instead the job of a teleprotection scheme, in which the relay at B sends a permissive signal to A confirming that the fault is on the line — which converts the delayed zone 2 into an instantaneous trip without changing any reach.

Review

Chapter Summary

Four qualities

Selectivity, speed, sensitivity, reliability — the last splitting into dependability and security.

Overlapping zones

Zones are bounded by CTs and overlap around each breaker so no point is unprotected.

Backup

Remote backup covers breaker failure; local backup is faster but shares the substation.

CT sizing

\(V_k \ge \mathrm{ALF}\times I_{sn}(R_{ct}+R_{lead}+R_b)\); never open a live CT secondary.

IDMT

\(t=\mathrm{TMS}\times 0.14/(M^{0.02}-1)\), with \(M=\) PSM \(=I_f/\)(CT ratio × plug setting).

Grading

Work from the load end back; check at the common fault; round TMS up; margin 0.3–0.4 s.

Direction

\(T\propto VI\cos(\theta-\tau)\) with the \(90^\circ\) connection; needed on rings and parallel feeders.

Percentage bias

\(I_{diff} > I_s + k I_{bias}\) — tolerance grows exactly where CT error grows.

Inrush

Blocked by second harmonic (15–20%); overexcitation by fifth. Bias alone cannot do it.

Distance zones

80% instantaneous, 120% at 0.35 s, line + 120% of longest adjacent at 0.8 s.

Infeed

\(Z_{app}=Z_{AB}+K Z_{Bx}\) — infeed under-reaches, and the weakest case sets the reach.

Numerical relays

Full-cycle DFT phasors, harmonics free, self-monitoring, IEC 61850 — same principles.

Practice

Practice Problems

Use the IEC standard-inverse characteristic \(t=\mathrm{TMS}\times 0.14/(M^{0.02}-1)\) unless another is named, and a grading margin of \(0.3\) s.

  1. A \(600/1\) A class \(5\text{P}20\) CT has a secondary resistance of \(5\) Ω and a pilot loop of \(1.5\) Ω. Find the knee-point voltage required for a \(0.2\) Ω numerical relay and for a \(4\) VA electromechanical relay. If the CT supplied has \(V_k = 150\) V, at what primary current does it saturate in the second case?
  2. A standard-inverse relay of \(5\) A rating is fed from a \(300/5\) CT with a plug setting of \(150\%\) and a TMS of \(0.4\). Find its operating time for a fault of \(3600\) A, and the time the same settings would give on a very-inverse characteristic. State which you would choose if the relay must discriminate against a downstream HRC fuse, and why.
  3. Relay P (CT \(200/1\), plug \(100\%\)) protects a feeder; relay Q (CT \(400/1\), plug \(100\%\)) is immediately upstream. A fault at the far end of P's section gives \(2400\) A. With P set to TMS \(0.10\), find its operating time and then the smallest TMS in steps of \(0.05\) that gives Q the required margin. State Q's actual margin.
  4. A directional relay has a maximum-torque angle of \(45^\circ\). For a forward fault the current lags the polarising voltage by \(70^\circ\). Find the torque as a fraction of its maximum value, and repeat for the same fault seen from the reverse direction. Explain why the \(90^\circ\) connection is used rather than feeding each element with its own phase voltage.
  5. A \(30\) MVA, \(66/11\) kV Dy11 transformer is to be given differential protection. Standard CT ratios of \(300/1\) on the HV side and \(1600/1\) on the LV side are available. Find the rated currents, the two CT secondary currents, and the percentage mismatch. State the minimum bias slope that mismatch alone demands, and how you would correct it.
  6. A differential relay has \(I_s = 0.2\) pu, a \(30\%\) slope to \(4\) pu of bias and \(60\%\) beyond, with the first breakpoint at \(1\) pu. Check its stability for a \(12\) pu through fault in which one CT reads \(8\%\) low, and find the largest CT error the setting would tolerate at that through current.
  7. A \(220\) kV line is \(120\) km long with \(z = 0.35\ \Omega\)/km. The shortest line leaving the remote bus is \(70\) km and the longest is \(150\) km, both with the same \(z\). With a \(600/1\) CT and a \(220\,000/110\) VT, set the three zones in primary and secondary ohms and state their time delays. Verify that zone 2 does not reach past the zone 1 of the shortest adjacent line.
  8. Explain why a full-cycle discrete Fourier transform rejects every harmonic exactly but does not reject the decaying DC offset, and what a mimic filter does about it. Then give one example of a change to a protection scheme that improves dependability while degrading security, and one that does the reverse.
Tip: in every settings problem, convert to primary amperes or primary ohms first and do the engineering there, then convert once at the end for the relay. The primary quantities are the ones you can sanity-check — a pickup must sit between full load and minimum fault current, a zone 1 reach must be shorter than the line, a bias threshold must sit above the tap-changer mismatch. A number that is wrong in primary terms is wrong; a number that looks strange in secondary terms is usually just a ratio you have not divided out yet.